3.3.29 \(\int \frac {\csc (a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [229]

Optimal. Leaf size=81 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}+\frac {2}{3 b d (d \cos (a+b x))^{3/2}} \]

[Out]

-arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)-arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)+2/3/b/d/(d*cos
(b*x+a))^(3/2)

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Rubi [A]
time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2645, 331, 335, 218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}+\frac {2}{3 b d (d \cos (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/(d*Cos[a + b*x])^(5/2),x]

[Out]

-(ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(5/2))) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(5/2)) + 2/(3
*b*d*(d*Cos[a + b*x])^(3/2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{5/2}} \, dx &=-\frac {\text {Subst}\left (\int \frac {1}{x^{5/2} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac {2}{3 b d (d \cos (a+b x))^{3/2}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d^3}\\ &=\frac {2}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^3}\\ &=\frac {2}{3 b d (d \cos (a+b x))^{3/2}}-\frac {\text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^2}-\frac {\text {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d^2}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{5/2}}+\frac {2}{3 b d (d \cos (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 70, normalized size = 0.86 \begin {gather*} -\frac {-2+3 \tan ^{-1}\left (\sqrt {\cos (a+b x)}\right ) \cos ^{\frac {3}{2}}(a+b x)+3 \tanh ^{-1}\left (\sqrt {\cos (a+b x)}\right ) \cos ^{\frac {3}{2}}(a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/(d*Cos[a + b*x])^(5/2),x]

[Out]

-1/3*(-2 + 3*ArcTan[Sqrt[Cos[a + b*x]]]*Cos[a + b*x]^(3/2) + 3*ArcTanh[Sqrt[Cos[a + b*x]]]*Cos[a + b*x]^(3/2))
/(b*d*(d*Cos[a + b*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(65)=130\).
time = 0.65, size = 624, normalized size = 7.70

method	result	size
default	\(\frac {\left (24 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {7}{2}}-12 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d^{3}-12 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d^{3}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-24 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {7}{2}}+12 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d^{3}+12 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d^{3}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {7}{2}}+4 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, d^{\frac {5}{2}}-3 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d^{3}-3 \sqrt {-d}\, \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d^{3}}{6 \sqrt {-d}\, d^{\frac {11}{2}} \left (4 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right ) b}\)	\(624\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*((24*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(7/2)-12*(-d)^(1/2)*ln(
2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*d^3-12*(-d)^(
1/2)*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*d^3)*
sin(1/2*b*x+1/2*a)^4+(-24*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(7/2)+
12*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-
d))*d^3+12*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x
+1/2*a)-d))*d^3)*sin(1/2*b*x+1/2*a)^2+6*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2
)-d))*d^(7/2)+4*(-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(5/2)-3*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1
)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))*d^3-3*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1
/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))*d^3)/(-d)^(1/2)/d^(11/2)/(4*s
in(1/2*b*x+1/2*a)^4-4*sin(1/2*b*x+1/2*a)^2+1)/b

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Maxima [A]
time = 0.60, size = 80, normalized size = 0.99 \begin {gather*} -\frac {\frac {6 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {3}{2}}} - \frac {3 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {3}{2}}} - \frac {4}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}}{6 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-1/6*(6*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(3/2) - 3*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x
+ a)) + sqrt(d)))/d^(3/2) - 4/(d*cos(b*x + a))^(3/2))/(b*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (65) = 130\).
time = 0.42, size = 318, normalized size = 3.93 \begin {gather*} \left [\frac {6 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{2} - 3 \, \sqrt {-d} \cos \left (b x + a\right )^{2} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{12 \, b d^{3} \cos \left (b x + a\right )^{2}}, -\frac {6 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{2} - 3 \, \sqrt {d} \cos \left (b x + a\right )^{2} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )}}{12 \, b d^{3} \cos \left (b x + a\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a)))*cos(b*x + a)^2
- 3*sqrt(-d)*cos(b*x + a)^2*log((d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) - 6*d*
cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a)))/(b*d^3*cos(b*x + a)^2), -1/
12*(6*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a)))*cos(b*x + a)^2 - 3*sq
rt(d)*cos(b*x + a)^2*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x +
a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) - 8*sqrt(d*cos(b*x + a)))/(b*d^3*cos(b*x + a)^2)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/(d*cos(b*x + a))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(5/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*cos(a + b*x))^(5/2)), x)

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